IIT JEE , AIEEE Forum - Physics , Mechanics

elessar_iitkgp

The length of the planck, L = 12 m

Initial velocity of the cylinder, ${v}_{0}$ = 7 m/s

Coefficient of friction between the cylinder and plank, $\mu$ = 0.1

Friction acts on the cylinder in a backwards direction in its full magnitude.

Equations:

$m{a}_{1} = \mu mg$

$\frac{1}{2} m{R}^{2} \alpha=\mu mgR$

$2m{a}_{2}=\mu mg$

Let at anytime t, the velocity of cylinder be ${v}_{1}$ , its angualr velocity be $\omega$ and the velocity of plank be ${v}_{2}$ . Let at time T the slipping is stopped. Then,

${v}_{1}(T)-R\omega(T)={v}_{2}(T)$

${v}_{0}-\mu gT-2\mu gT = (\mu g/2)T$

$T=(2{v}_{0})/(7\mu g)$

Substituting the values,

In these two seconds the cylinder has travelled the following distance on the plank

${L}_{1} = {v}_{0}T-(1/2){a}_{1,2}{T}^{2}$

where ${a}_{1,2}$ is the accelaration of the cylinder wrt the plank

Substituting values, ${L}_{1} = 11 m$

After the time , friction stops acting and the cylinder and the plank move with constant velocities

${v}_{1}={v}_{0}-\mu gT=5 m/s$

and

${v}_{2}=(\mu g/2)T = 1 m/s$

So time taken by the cylinder to cover the additional remaining is

$T'=(L-{L}_{1})/{v}_{1,2} = (1/4) s$

${v}_{1,2}$ being the relative velocity of the cylinder wrt the plank

So total time taken

Ask & Discuss Questions with Community & Experts